Primes and Irreducibles in Commutative Rings

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On this page:

• Introduction
• Primes and Irreducibles
• Factorization
• Worked Exercises

Introduction

In algebra, one learns that we can considerably generalize the notions of prime elements and a closely related notion of irreducibility. Let us start with notions of divisibility:

[definition] [deftitle]Definition (Divides, multiple, associates).[/deftitle]

Let $R$ be a commutative ring and let $x,y\in R$. We say that $x$ divides $y$ or that $y$ is a multiple of $x$ if there is some $r \in R$ for which

We write this as $x \mid y$. Equivalently, $y \in \generator{x}$. We say that $x$ and $y$ are associates if $\generator{x} = \generator{y}$. [/definition]

It is very tempting say that if $x$ and $y$ are associates, then $a = ub$ for some unit $u$ in $R$. This is NOT true in general. Accordingly, we work in the realm of IDs to get around pathological examples.

[theorem] [thmtitle]Proposition.[/thmtitle]

Let $R$ be an integral domain and $x, y \in R$ be nonzero. Then $x$ and $y$ are associates if and only if $x = uy$ for some unit $u\in R$. [/theorem]

[proof] ($\implies$) Since $\generator{x} = \generator{y} \neq 0$, there exists $r, s \in R$ for which

Thus, $x = rsx$ which implies that

and cancellative property of integral domains implies $rs = 1$ as $x$ and $y$ are necessarily nonzero. So $r$ and $s$ are units.

($\impliedby$) If $x = uy$, then $\generator{x} = \generator{uy} = \generator{y}$ and we are done. [/proof]

Primes and Irreducibles

[definition] [deftitle]Definition (Primes and irreducible elements).[/deftitle]

Let $R$ be an integral domain.

• We say that $x \in R$ is prime if $\generator{x}$ is prime.
• We say that $x \in R$ is irreducible if $x$ is nonunit and $x = rs$ implies that one of $r$ or $s$ is a unit in $R$. [/definition]

There are several different way to characterize irreducibility of nonzero irreducible elements:

[theorem] [thmtitle]Proposition.[/thmtitle]

Let $R$ be an integral domain and $x \in R$ be nonzero. Then the following are equivalent:

1. $x$ is irreducible.
2. $x = rs$ implies that $x$ is an associate of $r$ or of $s$.
3. $\generator{x} \subseteq \generator{y}$ implies $\generator{y} = \generator{x}$ or $\generator{y} = R$.
4. $\generator{x}$ is maximal among proper principal ideals. [/theorem]

[proof] (1) $\implies$ (2). If $x$ is irreducible, one of $r$ and $s$ is a unit. By the proposition above, $x$ is an associate of one of $r$ or $s$.

(2) $\implies$ (3). $\generator{x} \subseteq \generator{y}$ implies that there is some $r \in R$ for which $x = ry$. So $x$ is an associate of $r$ or of $y$ by (2). If $x$ is an associate of $y$, then $\generator{x} = \generator{y}$. If $x$ is an associate of $r$, then the proposition above implies that $x = ur$ for some unit $u \in R$. Thus, $ur = ry$. As $x$ is nonzero, $r$ is also nonzero so cancellation implies $u = y$. So $y$ is a unit and, thus, $\generator{y} = R$.

(3) $\implies$ (4). Immediate.

(4) $\implies$ (1). Since $\generator{x}$ is a proper ideal, $x$ is not a unit. Now say $x = rs$ for some $r, s \in R$. The maximality condition on $\generator{x}$ implies that $\generator{x} = (r)$ or $\generator{x} = \generator{s}$. Without loss of generality, say $\generator{x} = \generator{r}$. Then $x$ is an associate of $r$ which implies that $x = ur$ for some unit $u \in R$. But this implies that $ur = rs$ and the cancellative property of IDs implies that $u = s$. So $s$ is a unit and we are done. [/proof]

A tempting thing to say is that primes and irreducibles are equivalent. This is NOT true in general. In particular, irreducibles can fail to be the prime. The converse is always true, though.

[theorem] [thmtitle]Proposition (Primes are always irreducible).[/thmtitle]

Let $R$ be an integral domain and let $x \in R$ be a nonzero prime. Then $x$ is irreducible. [/theorem]

[proof] Let $x = rs$. Then either $x \mid r$ or $x \mid s$. Without loss of generality, say $x \mid r$. Then $r = tx$ for some $t \in R$. So then

and the cancellative property says that $st = 1$. Thus, $s$ and $t$ are units in $R$. So $x$ is irreducible as desired. [/proof]

[example] [extitle]An irreducible that is not prime in $R \coloneqq \ZZ[\sqrt{-5}]$.[/extitle]

Define the norm function $N:R \to \ZZ_{\ge 0}$ by

With a little bit of high-school algebra, we can show that

We claim that this construction can be used to show that $3$ is irreducible yet not prime in $R$. To show irreducibility, If can be written as

then

Noting that. Since $N(x)N(y)$ are nonnegative integer factors of $9$, we know that we must have (up to WLOG) $N(x) = 1$ and $N(y) = 9$, or $N(x) = 3$ and $N(y) = 3$. Now, if $N(x) = 1$, then $x = \pm 1$ so $x$ is a unit. On the other hand, if $N(x) = 3$, then

and this obviously has no solutions in $\ZZ$ (let alone $\ZZ_{\ge 0}$). Thus, one of $x$ and $y$ must be a unit and we have shown irreducibility.

To show that $3$ is not prime, we proceed by contradiction. Assume that $3$ is prime and consider that

Since $3\mid 3^2$, it follows that $3$ divides one of the factors on the right. If $3 \mid (2 + \sqrt{-5})$m then

for some $a, b \in \ZZ$. So we have $3b = 1$ but this is impossible. An identical argument shows that $3 \not\mid (2 - \sqrt{-5})$ and, so, $3$ is not prime. [/example]

It is easy to see that $\ZZ[\sqrt{-5}]$ is an ID. So notice that even an ID can lead to counterintuitive results when it comes to factoring. This is why we are interested in stronger notions of a domain such as that of UFDs, PIDs, Euclidean domains, etc.

Factorization

We model the idea of factorization on the same idea of prime factorization in $\ZZ$. Since we just saw that primes are a much stronger notion than irreducibles, it is reasonable to question whether or not (nonzero) prime elements actually exist. Hence, we talk about the more controllable irreducibles instead:

[definition] [deftitle]Definition (Factorization).[/deftitle]

Let $R$ be an integral domain. We say that $r \in R$ has a factorization into irreducibles if we can write $r$ in the form

where each $q_i$ is an irreducible element of $R$. We say this factorization is unique if the elements are determined by $r$ up to order and associates. [/definition]

Obvious examples are polynomial factorizations and integer factorizations. Less trivial factorizations include the Gaussian integers.

As it turns out (should be no surprise), if $r$ has a factorization into irreducibles, then we can characterize that quality via ACCs. In particular, we must be able to say that factorizations (possibly into reducibles) must eventually terminate at irreducibles.

[theorem] [thmtitle]Proposition (ACC characterization of factorization into irreducibles).[/thmtitle]

Let $R$ be an integral domain and $r \in R$ be nonzero and nonunit. If every ACC of principal ideals

stabilizes, then $r$ has a factorization into irreducibles. [/theorem]

[proof] For the sake of contradiction, suppose that $r$ fails to have a factorization into irreducibles. Then, that means we can factor $r$ into

Now, one of $r_1$ and $x_1$ need to be reducible. Without loss of generality, say $r_1$ is reducible. Then we can factor $r_1$ into

Same argument allows us to say, without loss of generality, that $r_2$ is reducible. Continuing this process inductively, we construct an AC

That does NOT stabilize. [/proof]

[example] [extitle]Remark.[/extitle]

Again, notice that we need to ensure EVERY said AC stabilizes. Not just one of them. The irreducibility condition is not strong enough to force the irreducible to divide at least one of the factors in the way a prime element can. An easy counterexample is to consider

Clearly this stabilizes but $r$ can easily fail to be irreducible. [/example]

There is a particularly obvious case when factorizations exist.

[theorem] [thmtitle]Corollary.[/thmtitle]

Let $R$ be an integral domain that is also Noetherian (i.e. a Noetherian domain). Then factorizations exist in $R$. [/theorem]

[proof] In Noetherian rings, the ACC ensures that every AC of principal ideals stabilizes. Thus, the claim follows by the above. [/proof]

Finally, we point out that a special class of rings that are particularly important that we will come back to in another page.

[definition] [deftitle]Definition (UFDs).[/deftitle]

Let $R$ be an integral domain. We say that $R$ is a domain with factorizations if every nonzero, nonunit element has a factorization into irreducibles. If it is also the case that said factorization is unique, then we say that $R$ is a unique factorization domain (UFD). [/definition]

[example] [extitle]$\ZZ[\sqrt{-5}]$ is NOT a UFD.[/extitle]

Recall that

We showed that $3$ is irreducible. It’s not difficult to also show that $2 \pm \sqrt{-5}$ is as well by using the multiplicative property of norm. [/example]

[example] [extitle]Domain with factorizations $\not\Rightarrow$ Noetherian.[/extitle]

Not every domain with factorizations is Noetherian. For instance, the ring

clearly has factorizations but it is also NOT Noetherian. It is straightforward to see that the AC

does not stabilize. [/example]

Worked Exercises